320 



OF THE POSITIONS OP EQUILIBRIUM. 



gravity of the solid 0.565 as in the preceding example ; how much of 

 the equal sides is immersed in the fluid, and how much is extant, the 

 body being in a state of quiescence ? 



Here, by the rule, we have 



ls= 1 0.565 = 0.435, 



the square root of which is 



v/ 0.435 zz 0.659; 



consequently, by multiplication, we have 

 *or2/zz28x0.659:z: 18.452 inches. 



404. And the position of equilibrium indicated by the above value 

 of x and y, is as represented in the subjoined 



diagram, where IK is the horizontal surface 

 of the fluid, ABED the immersed portion of 

 the section, and DCE the extant portion, DE 

 being the water line or plane of floatation. 



Since CD and CE are each equal to 18.452 

 inches, it follows, that AD and BE are each 

 equal to 28 18.452 zz 9.548 inches, the 

 extant part of the equal sides being nearly 

 double of the immersed part. 



Bisect A B in the point F, and draw the straight lines F D and F E ; 

 then shall FD and FE be equal to one another; this is manifest, for 

 AF, BF and AD, BE are equal, and the angle DAF is equal to the 

 angle EBF ; therefore, FD is equal to FE. 



By examining the nature of the equation (243) or (244), it is 

 manifest that the values of x and y depend entirely on the value of s, 

 or the specific gravity of the floating body ; now, since this may admit 

 of all magnitudes between zero and unity, which is the specific gravity 

 of water, it follows, that x and y may be of all magnitudes between 

 zero and 28 inches ; but whatever may be the magnitude of the extant 

 sides, the position in which the body floats will be the same ; viz. that 

 in which the base of the section is parallel to the surface of the fluid. 



405. Admitting the specific gravity of the solid to fall within the 

 limits of possibility, the formula equation (242), when reduced, will 

 supply us with other two positions of equilibrium, in which the body 

 may float with two of its angles immersed ; here follows the reduction 

 of the equation. 



Complete the square, and we obtain 



2 a 2 ) b\\ s), 



